Question: $\dfrac{ 9e + 4f }{ -3 } = \dfrac{ 8e - 4g }{ -8 }$ Solve for $e$.
Answer: Multiply both sides by the left denominator. $\dfrac{ 9e + 4f }{ -{3} } = \dfrac{ 8e - 4g }{ -8 }$ $-{3} \cdot \dfrac{ 9e + 4f }{ -{3} } = -{3} \cdot \dfrac{ 8e - 4g }{ -8 }$ $9e + 4f = -{3} \cdot \dfrac { 8e - 4g }{ -8 }$ Multiply both sides by the right denominator. $9e + 4f = -3 \cdot \dfrac{ 8e - 4g }{ -{8} }$ $-{8} \cdot \left( 9e + 4f \right) = -{8} \cdot -3 \cdot \dfrac{ 8e - 4g }{ -{8} }$ $-{8} \cdot \left( 9e + 4f \right) = -3 \cdot \left( 8e - 4g \right)$ Distribute both sides $-{8} \cdot \left( 9e + 4f \right) = -{3} \cdot \left( 8e - 4g \right)$ $-{72}e - {32}f = -{24}e + {12}g$ Combine $e$ terms on the left. $-{72e} - 32f = -{24e} + 12g$ $-{48e} - 32f = 12g$ Move the $f$ term to the right. $-48e - {32f} = 12g$ $-48e = 12g + {32f}$ Isolate $e$ by dividing both sides by its coefficient. $-{48}e = 12g + 32f$ $e = \dfrac{ 12g + 32f }{ -{48} }$ All of these terms are divisible by $4$ Divide by the common factor and swap signs so the denominator isn't negative. $e = \dfrac{ -{3}g - {8}f }{ {12} }$